Problem: Let $
f(n) =
\begin{cases}
n^2+1 & \text{if }n\text{ is odd} \\
\dfrac{n}{2} & \text{if }n\text{ is even}
\end{cases}.
$

For how many integers $n$ from 1 to 100, inclusive, does $f ( f (\dotsb f (n) \dotsb )) = 1$ for some number of applications of $f$?
First, we note that if $n$ is a positive integer, then $f(n)$ is also a positive integer.  We claim that $f ( f (\dotsb f (n) \dotsb )) = 1$ for some number of applications of $f$ only for $n = 1, 2, 4, 8, 16, 32,$ and $64.$  (In other words, $n$ must be a power of 2.)

Note that $f(1) = 2,$ so $f(f(1)) = f(2) = 1.$  If $n > 1$ is a power of 2, it is easy to see that repeated applications of $f$ on $n$ eventually reach 1.

Suppose $n$ is an odd positive integer, where $n > 1.$  Write $n = 2k + 1,$ where $k$ is a positive integer.  Since $n$ is odd,
\[f(n) = n^2 + 1 = (2k + 1)^2 + 1 = 4k^2 + 4k + 2 = 2(2k^2 + 2k + 1).\]Since $2k^2 + 2k$ is always even, $2k^2 + 2k + 1$ is always odd (and greater than 1), so $f(n)$ can never be a power of 2 when $n$ is odd and greater than 1.

Now, suppose $n$ is even.  For example, if $n = 2^3 \cdot 11,$ then
\[f(2^3 \cdot 11) = f(2^2 \cdot 11) = f(2 \cdot 11) = f(11),\]which we know is not a power of 2.

More generally, suppose $n = 2^e \cdot m,$ where $e$ is nonnegative and $m$ is odd.  Then
\[f(2^e \cdot m) = f(2^{e - 1} \cdot m) = f(2^{e - 2} \cdot m) = \dots = f(m).\]If $m = 1,$ then $n$ is a power of 2, and the sequence eventually reaches 1.  Otherwise, $f(m)$ is not a power of 2.  We also know that $f(m)$ is odd and greater than 1, $f(f(m))$ is not a power of 2 either, and so on.  Thus, the sequence can never reach 1.

Therefore, $n$ must be one of the $\boxed{7}$ values 1, 2, 4, 8, 16, 32, or 64.